Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A car is negotiating a curved road of radius $R$ . The road is banked at an angle $\theta$ . The coefficient of friction between the tyre of the car and the road is $\mu_s$ . The maximum safe velocity on this road is:

$\sqrt{gR \left( \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right)}$

$\sqrt{\frac{g}{R} \left( \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right)}$

$\sqrt{gR^2 \left( \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right)}$

$\sqrt{gR^2 \left( \frac{\mu_s + \tan \theta}{1 + \mu_s \tan \theta} \right)}$

Step-by-Step Solution

Forces Analysis: Consider the car as a particle. The forces acting on it are: weight $mg$ (downwards), normal reaction $N$ (perpendicular to the road), and static friction $f$ (parallel to the road). For maximum speed ( $v_{max}$ ), friction acts down the slope to prevent sliding up.
Equations of Motion:

Vertical equilibrium: $N \cos \theta = mg + f \sin \theta$ . Since $f = \mu_s N$ (limiting case), $N \cos \theta - \mu_s N \sin \theta = mg$ .
Horizontal centripetal force: $N \sin \theta + f \cos \theta = \frac{mv^2}{R}$ . Substituting $f = \mu_s N$ , $N \sin \theta + \mu_s N \cos \theta = \frac{mv_{max}^2}{R}$ .

Derivation: Divide the horizontal equation by the vertical equation: $\frac{\frac{mv_{max}^2}{R}}{mg} = \frac{N(\sin \theta + \mu_s \cos \theta)}{N(\cos \theta - \mu_s \sin \theta)}$ $\frac{v_{max}^2}{Rg} = \frac{\sin \theta + \mu_s \cos \theta}{\cos \theta - \mu_s \sin \theta}$ Divide the numerator and denominator of the RHS by $\cos \theta$ : $\frac{v_{max}^2}{Rg} = \frac{\tan \theta + \mu_s}{1 - \mu_s \tan \theta}$ $v_{max} = \sqrt{Rg \left( \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right)}$ (Reference: NCERT Class 11, Physics Part I, Chapter 5, Section 5.10, Equation 4.21).

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